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Q: If using numbers 1 thru 49 how many 6 digit combinations can be made?

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120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.

66

15

10,000

2304

9^9 = 387420489 different digit combinations.

If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.

If each can be used again, it is 7x7x7x7 or 2401 combinations. If not it is 7x6x5x4 or 840 combinations

It is: 9C7 = 36

Through the magic of perms and coms the answer is 729

Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.

In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.

10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).

10,000 Because it begins with 0000, 0001, 0002... and goes to 9999.

There are 9999 possible combinations starting from 0000 to 9999

6 ways: 931,913,139,193,391,319

the first digit can be any of the seven numbers, the second can be any of the remaining 6 etc so there are 7 x 6 x 5 x 4 x 3 x 2 x 1 combinations. This is known as "factorial 7" and is normally written "7!" and is 5040.

9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.

a lot

The answer is 10C4 = 10!/[4!*6!] = 210

If the numbers are not allowed to repeat themselves then 362,880. * * * * * That is the number of permutations, not combinations. In a combination the order of the digits does not matter. So there is only one 9-digit combination.

The right answer is 2763633600 by amar nirania

210

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